Problem: We know that $-\frac{4}{1+16x^2}=-4+64x^2-1024x^4+16384x^6-...$ for $x\in\left(-\dfrac{1}{4},\dfrac{1}{4}\right)$. Using this fact, find the function that corresponds to the following series. $ -4x+\frac{{64}}{3}x^3-\frac{1024}{5}x^5+\frac{16384}{7}x^7-...$ Choose 1 answer: Choose 1 answer: (Choice A) A $\arctan(4x)$ (Choice B) B $\arctan(-4x)$ (Choice C) C $-\arctan(-4x)$ (Choice D) D $\ln(-4x)$ (Choice E) E $-\ln(4x)$ (Choice F) F $\ln(4x)$
Explanation: First, note that the derivative of $ -4x+\frac{{64}}{3}x^3-\frac{1024}{5}x^5+\frac{16384}{7}x^7-...$ is $ -4+64x^2-1024x^4+16384x^6-...=-\frac{4}{1+16x^2}$ Then $\int{\left(-\frac{4}{1+16x^2}\right)}dx=\int\left(-4+64x^2-1024x^4+16384x^6-...\right)dx$ Since $\int{-\frac{4}{1+16x^2}}~dx$ can be rewritten as $\int{\frac{-4}{1+(-4x)^2}}~dx$, we see that $\int{-\frac{4}{1+16x^2}}~dx=\arctan(-4x)+C$ Thus $\arctan(-4x) + C=-4x+\frac{{64}}{3}x^3-\frac{1024}{5}x^5+\frac{16384}{7}x^7-...$ Now let $x=0$ ; since $\arctan(0)=0$, we know that $C=0$. Hence, $ -4x+\frac{{64}}{3}x^3-\frac{1024}{5}x^5+\frac{16384}{7}x^7-...=\arctan(-4x)$